Sturm separation theorem

In mathematics, in the field of ordinary differential equations, Sturm separation theorem, named after Jacques Charles François Sturm, describes the location of roots of homogeneous second order linear differential equations. Basically the theorem states that given two linear independent solution of such an equations the roots of the two solutions are alternating.

Sturm separation theorem

Given a homogeneous second order linear differential equation and two continuous linear independent solutions "u"("x") and "v"("x") with "x"₀ and "x"₁ successive roots of "u"("x"), then "v"("x") has exactly one root in the open interval ] "x"₀, "x"₁ [.

Proof

The proof is by contradiction. Assume that "v" has no zeros in ] "x"₀, "x"₁ [. Since "u" and "v" are linearly independent, "v" cannot vanish at either "x"₀ or "x"₁, so the quotient "u" / "v" is well-defined on the closed interval ["x"₀, "x"₁] , and it is zero at "x"₀ and "x"₁. Hence, by Rolle's theorem, there is a point ξ between "x"₀ and "x"₁ where :$$frac{d}{dx} left( frac{u(x)}{v(x)} ight) = frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2} vanishes. Hence, $$u'(xi) v(xi) = u(xi) v'(xi), which implies that "u" and "v" are linearly dependent. This contradicts our assumption, and thus, "v" has to have at least one zero between "x"₀ and "x"₁.

On the other hand, there can be only one zero between "x"₀ and "x"₁, because otherwise "v" would have two zeros and there would be no zeros of "u" in between, and it was just proved that this is impossible harv|Bhamra|Ratna Bala|2003|p=262.

An Alternative Proof

Since $$displaystyle u and $$displaystyle v are linearly independent it follows that the Wronskian $$displaystyle W [u,v] must satisfy $$W [u,v] (x)equiv W(x) eq 0 for all $$displaystyle x where the differential equation is defined, say $$displaystyle I. Without loss of generality, suppose that $$W(x)<0mbox{ }forallmbox{ }xin I. Then:$$u(x)v'(x)-u'(x)v(x) eq 0So at $$displaystyle x=x_0:$$W(x)=-u'left(x_0 ight)vleft(x_0 ight)and either $$u'left(x_0 ight) and $$vleft(x_0 ight) are both positive or both negative. Without loss of generality, suppose that they are both positive. Now, at $$displaystyle x=x_1 :$$W(x)=-u'left(x_1 ight)vleft(x_1 ight)and since $$displaystyle x=x_0 and $$displaystyle x=x_1 are successive zeros of $$displaystyle u(x) it causes $$u'left(x_1 ight)<0. Thus, to keep $$displaystyle W(x)<0 we must have $$vleft(x_1 ight)<0. We see this by observing that if $$displaystyle u'(x)>0mbox{ }forallmbox{ }xin left(x_0,x_1 ight] then $$displaystyle u(x) would be increasing (away from the $$displaystyle x-axis), which would never lead to a zero at $$displaystyle x=x_1. So for a zero to occur at $$displaystyle x=x_1 at most $$u'left(x_1 ight)=0 (i.e., $$u'left(x_1 ight)leq 0 and it turns out, by our result from the Wronskian that $$u'left(x_1 ight)leq 0). So somewhere in the interval $$left(x_0,x_1 ight) $$displaystyle v(x) changed signs. By the Intermediate Value Theorem there exists $$x^*inleft(x_0,x_1 ight) such that $$vleft(x^* ight)=0.

By the same reasoning as in the first proof, $$displaystyle v(x) can have at most one zero for $$xin left(x_0,x_1 ight).

References

*.